Field if I define The plane

Bachelier

if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) | a, b ∈ ℝ \} ##

and define addition and multiplication as:
(a, b) + (c, d) := (a + c, b + d)
(a, b) · (c, d) := (ac, bd)

Then ##F## is a field. right?

would the multiplication as described here make ℂ a field?

pwsnafu

Quote by Bachelier

if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) | a, b ∈ ℝ \} ##

and define addition and multiplication as:
(a, b) + (c, d) := (a + c, b + d)
(a, b) · (c, d) := (ac, bd)

Then ##F## is a field. right?

No, you have zero divisors.

would the multiplication as described here make ℂ a field?

How would you do i² = -1?

Bachelier

Quote by pwsnafu

No, you have zero divisors.

You mean like (0,7).(8,0)

HallsofIvy

Yes, neither (7, 0) nor (0, 8) is the additive identity but neither has a mulitplicative inverse.

Bachelier

thanks. Basically ℂ will fail to be an integral domain in the first place under this operation.

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Bachelier	Field if I define The plane if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) \| a, b ∈ ℝ \} ## and define addition and multiplication as: (a, b) + (c, d) := (a + c, b + d) (a, b) · (c, d) := (ac, bd) Then ##F## is a field. right? would the multiplication as described here make ℂ a field?

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	Field if I define The plane Yes, neither (7, 0) nor (0, 8) is the additive identity but neither has a mulitplicative inverse.